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Home/ Questions/Q 718721
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T05:33:35+00:00 2026-05-14T05:33:35+00:00

How do I remove the oldItems from the newItems string? (remove Blue and Green)

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How do I remove the oldItems from the newItems string? (remove Blue and Green) newItems is generated from jQuery autocomplete and I would like to remove already selected items from the select list.

newItems = Blue \n Red \n Black \n Yellow \n Green \n
oldItems = Blue,Yellow,Orange,Green

Best regards.
Asbjørn Morell.

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  1. Editorial Team
    Editorial Team
    2026-05-14T05:33:35+00:00Added an answer on May 14, 2026 at 5:33 am

    One algorithm would be to look at each of the oldItems and search for them in the newItems array. However, if you expect these arrays to reach any length, this is going to be O(n^2). This is essentially the algorithm given by Jacob Relkin.

    However, instead, if you sort the two lists, you can do it faster.

    var newArray = newItems.split("\n").sort();
    var oldArray = oldItems.split(",").sort();
    var newUniq = [];
    
    var newLength = newArray.length;
    var oldLength = oldArray.length;
    
    var oldI = 0;
    var newI = 0;
    while (newI < newLength && oldI < oldLength) {
       var oldString = oldArray[oldI].trim();
       var newString = newArray[newI].trim();
       if (oldString == "") {
          oldI++;
       } else if (newString == "") {
          newI++;
       } else if (oldString == newString) {
          /* We only update newI, because if there are multiple copies of the 
           * same string in newItems, we want to ignore them all. */
          newI++;
       } else if (oldString < newString) {
          oldI++;
       } else { /* newArray[newI] < oldArray[oldI] */
          newUniq.push(newArray[newI]);
          newI++;
       }
    }
    while (newI < newLength) {
       newUniq.push(newArray[newI]);
       newI++;
    }
    
    newItems = newUniq.join(" \n ");
    

    This walks through the two lists looking for duplicates in the new list in the old list. This should be O(n log n) because of the two sort operations.

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